25 Differentiation and the Laplace Transform In this chapter, we explore how the Laplace transform interacts with the basic operators of calculus: differentiation and integration. The greatest interest will be in the first identity that we will derive. This relates the transform of a derivative of a function to the transform of the original function, and will allow us to convert many initial-value problems to easily solved algebraic equations. But there are other useful relations involving the Laplace transform and either differentiation or integration. So we’ll look at them, too. 25.1 Transforms of Derivatives The Main Identity To see how the Laplace transform can convert a differential equation to a simple algebraic equation, let us examine how the transform of a function’s derivative, d f ∞ d f ∞ d f L f ′(t) = L = e−st dt = e−st dt , s dt dt dt s Z0 Z0 is related to the corresponding transform of the original function, ∞ −st F(s) = L[ f (t)]|s = f (t)e dt . Z0 The last formula above for L f ′(t) clearly suggests using integration by parts, and to ensure that this integration by parts is valid, we need to assume f is continuous on [0, ∞) and f ′ is at least piecewise continuous on (0, ∞) . Assuming this, ∞ d f L f ′(t) = e−st dt s dt Z0 u dv ∞ ∞|{z} | {z } = uv t=0 − v du Z0 ∞ −st ∞ −st = e f (t) t=0 − f (t) − se dt Z0 507 508 Differentiation and the Laplace Transform ∞ = lim e−st f (t) − e−s·0 f (0) − − se−st f (t) dt t→∞ Z0 ∞ = lim e−st f (t) − f (0) + s f (t)e−st dt . t→∞ Z0 Now, if f is of exponential order s0 , then −st lim e f (t) = 0 whenever s > s0 t→∞ and ∞ −st F(s) = L[ f (t)]|s = f (t)e dt exists for s > s0 . Z0 ′ Thus, continuing the above computations for L f (t) with s > s0 , we find that ∞ L f ′(t) = lim e−st f (t) − f (0) + s f (t)e−st dt s t→∞ Z0 = 0 − f (0) + sL[ f (t)]|s , which is a little more conveniently written as L ′ L f (t) s = s [ f (t)]|s − f (0) (25.1a) or even as L ′ f (t) s = s F(s) − f (0) . (25.1b) This will be a very useful result, well worth preserving in a theorem. Theorem 25.1 (transform of a derivative) ′ Let F = L[ f ] where f is a continuous function of exponential order s0 on [0, ∞) . If f is at least piecewise continuous on (0, ∞) , then L ′ f (t) s = s F(s) − f (0) for s > s0 . Extending these identities to formulas for the transforms of higher derivatives is easy. First, for convenience, rewrite equation (25.1a) as L ′ L g (t) s = s [g(t)]|s − g(0) or, equivalently, as dg L = sL[g(t)]| − g(0) . dt s s ′ (Keep in mind that this assumes g is a continuous function of exponential order, g is piecewise continuous and s islargerthantheorderof g .) Nowwesimplyapplythisequationwith g = f ′ , g = f ′′ , etc. Assuming all the functions are sufficiently continuous and are of exponential order, we see that d f ′ L f ′′(t) = L = sL f ′(t) − f ′(0) s dt s s ′ = s [s F(s) − f (0)] − f (0) = s2 F(s) − sf (0) − f ′(0) . Transforms of Derivatives 509 Using this, we then see that d f ′′ L f ′′′(t) = L = sL f ′′(t) − f ′′(0) s dt s s = s s2 F(s) − sf (0) − f ′(0) − f ′′(0) = s3 F(s) − s2 f (0) − sf ′(0) − f ′′(0) . Clearly, if we continue, we will end up with the following corollary to theorem 25.1: Corollary 25.2 (transforms of derivatives) ′ Let F = L[ f ] where f is a continuous function of exponential order s0 on [0, ∞) . If f is at least piecewise continuous on (0, ∞) , then L ′ f (t) s = s F(s) − f (0) for s > s0 . ′ ′′ If, in addition, f is a continuous function of exponential order s0 , and f is at least piecewise continuous, then L ′′ 2 ′ f (t) s = s F(s) − sf (0) − f (0) for s > s0 . ′ ′′ (n−1) More generally, if f , f , f , …, and f are all continuous functions of exponential order (n) s0 on [0, ∞) for some positive integer n , and f is at least piecewise continuous on (0, ∞) , then, for s > s0 , L (n) n n−1 n−2 ′ f (t) s = s F(s) − s f (0) − s f (0) − sn−3 f ′′(0) − ··· − sf (n−2)(0) − f (n−1)(0) . Using the Main Identity Let us now see how these identities can be used in solving initial-value problems. We’ll start with something simple: !◮Example 25.1: Consider the initial-value problem dy − 3y = 0 with y(0) = 4 . dt Observe what happens when we take the Laplace transform of the differential equation (i.e., we take the transform of both sides). Initially, we just have dy L − 3y = L[0]| . dt s s By the linearity of the transform and fact that L[0] = 0 , this is the same as dy L − 3L[y]| = 0 . dt s s 510 Differentiation and the Laplace Transform Letting Y = L[y] and applying the “transform of the derivative identity” (theorem 25.1, above), our equation becomes sY (s) − y(0) − 3Y (s) = 0 , which, since the initial condition is y(0) =4 , can be rewritten as sY (s) − 4 − 3Y (s) = 0 . This is a simple algebraic equation that we can easily solve for Y (s) . First, gather the Y (s) terms together and add 4 to both sides, [s − 3]Y (s) = 4 , and then divide through by s − 3 , 4 Y (s) = . s − 3 Thus, we have the Laplace transform Y of the solution y to the original initial-value problem. Of course, it would be nice if we can recover the formula for y(t) from Y (s) . And this is fairly easy done, provided we remember that 4 1 = 4 · = 4 L e3t = L 4e3t . s − 3 s − 3 s s Combining the last two equations with the definition of Y , we now have 4 L[y(t)]| = Y (s) = = L 4e3t . s s − 3 s That is, L[y(t)] = L 4e3t , from which it seems reasonable to expect y(t) = 4e3t . We will confirm that this is valid reasoning when we discuss the “inverse Laplace transform” in the next chapter. In general, it is fairly easy to find the Laplace transform of the solution to an initial-value problem involving a linear differential equation with constant coefficients and a ‘reasonable’ forcing function1. Simply take the transform of both sides of the differential equation involved, apply the basic identities, avoid getting lost in the bookkeeping, and solve the resulting simple algebraic equation for the unknown function of s . But keep in mind that this is just the Laplace transform Y (s) of the solution y(t) to the original problem. Recovering y(t) from the Y (s) found will usually not be as simple as in the last example. We’ll discuss this (the recovering of y(t) from Y (s) ) in greater detail in the next chapter. For now, let us just practice finding the “ Y (s) ”. 1 i.e., a forcing function whose transform is easily computed Transforms of Derivatives 511 ◮ ! Example 25.2: Let’s find the Laplace transform Y (s) = L[y(t)]|s when y is the solution to the initial-value problem y′′ − 7y′ + 12y = 16e2t with y(0) = 6 and y′(0) = 4 . Taking the transform of the equation and proceeding as in the last example: L ′′ ′ L 2t y − 7y + 12y s = 16e s L ′′ L ′ L L 2t y s − 7 y s + 12 [y]|s = 16 e s →֒ (s2Y (s) − sy(0) − y′(0 →֒ 16 − 7 sY (s) − y(0) + 12Y (s) = s − 2 s2Y (s) − s6 − 4 →֒ 16 − 7 sY (s) − 6 + 12Y (s) = s − 2 s2Y (s) − 6s − 4 →֒ 16 − 7sY (s) + 7 · 6 + 12Y (s) = s − 2 16 = s2 − 7s + 12 Y (s) − 6s + 38 →֒ s − 2 16 . s2 − 7s + 12 Y (s) = + 6s − 38 →֒ s − 2 Thus, 16 6s − 38 Y (s) = + . (25.2) (s − 2)(s2 − 7s + 12) s2 − 7s + 12 If desired, we can obtain a slightly more concise expression for Y (s) by finding the common denominator and adding the two terms on the right, 16 (s − 2)(6s − 38) Y (s) = + , (s − 2)(s2 − 7s + 12) (s − 2) s2 − 7s + 12 obtaining 6s2 − 50s + 92 Y (s) = . (25.3) (s − 2) s2 − 7s + 12 We will finish solving the above initial-value problem in exa mple 26.6 on page 533. At that time, we will find the later expression for Y (s) to be more convenient. At this point, though, there is no significant advantage gained by reducing expression (25.2) to (25.3). When doing similar problems in the exercises, go ahead and “find the common denominator and add” if the algebra is relatively simple. Otherwise, leave your answers as the sum of two terms. However, do observe that we did NOT multiply out the factors in the denominator, but left them as (s − 2) s2 − 7s + 12 . 512 Differentiation and the Laplace Transform Do the same in your own work. In the next chapter, we will see that leaving the denominator in factored form will simplify the task of recovering y(t) from Y (s) .